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Difference between revisions of "1995 AIME Problems/Problem 12"

(incomplete; both solutions written by 4everwise)
(finishing)
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=== Solution 1 (trigonometry) ===
 
=== Solution 1 (trigonometry) ===
 
<center><asy>
 
<center><asy>
 +
size(220); defaultpen(linewidth(0.7)); currentprojection = perspective(5,3,2);
 
import three;
 
import three;
−
triple A = (1,0,0), B=(0,0,0), C=(0,1,0), D=(1,1,0), O=(1,1,(1+2^.5)^.5)/2^.5, P=O*(18^.5-2)/5; /* , P = foot(A, O, B) */
+
triple A = (1,0,0), B=(0,0,0), C=(0,1,0), D=(1,1,0), O=(1,1,(1+2^.5)^.5)/2^.5, P=O*(1-.5^.5); /* , P = foot(A, O, B) */
−
draw(A--B--C--D--A--O--B--O--C--O--D); D(A--P--C);
+
draw(A--B--C--D--A--O--B--O--C--O--D); draw(A--C,linewidth(0.6)); D(A--P--C); MP("A",A);MP("B",B);MP("C",C);MP("D",D);MP("P",P,NW);MP("\theta",P,(0.5,-4));MP("45^{\circ}",O,(2,-6)); MP("O",O,N);
 +
draw(rightanglemark(C,P,O,2)); draw(anglemark(A,P,C,3)); draw(anglemark(D,O,C,3));
 
</asy></center>
 
</asy></center>
  
−
{{incomplete|Asymptote}}
+
The angle <math>\theta</math> is the angle formed by two [[perpendicular]]s drawn to <math>BO</math>, one on the plane determined by <math>OAB</math> and the other by <math>OBC</math>. Let the perpendiculars from <math>A</math> and <math>C</math> to <math>\overline{OB}</math> meet <math>\overline{OB}</math> at <math>P.</math> [[Without loss of generality]], let <math>AP = 1.</math> It follows that <math>\triangle OPA</math> is a <math>45-45-90</math> [[right triangle]], so <math>OP = AP = 1,</math> <math>OB = OA = \sqrt {2},</math> and <math>AB = \sqrt {4 - 2\sqrt {2}}.</math> Therefore, <math>AC = \sqrt {8 - 4\sqrt {2}}.</math>
−
 
 
−
The angle <math>\theta</math> is the angle formed by two [[perpendicular]]s drawn to <math>BO</math>, one on the plane determined by <math>OAB</math> and the other by <math>OBC</math>. Let the perpendiculars from <math>A</math> and <math>C</math> to <math>\overline{OB}</math> meet <math>\overline{OB}</math> at <math>P.</math> [[Without loss of generality]], let <math>AP = 1.</math> It follows that <math>OP = AP = 1,</math> <math>OB = OA = \sqrt {2},</math> and <math>AB = \sqrt {4 - 2\sqrt {2}}.</math> Therefore, <math>AC = \sqrt {8 - 4\sqrt {2}}.</math>
 
  
 
From the [[Law of Cosines]], <math>AC^{2} = AP^{2} + PC^{2} - 2(AP)(PC)\cos \theta,</math> so
 
From the [[Law of Cosines]], <math>AC^{2} = AP^{2} + PC^{2} - 2(AP)(PC)\cos \theta,</math> so
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Thus <math>m + n = \boxed{005}</math>.
 
Thus <math>m + n = \boxed{005}</math>.
  
−
=== Solution 2 (analytic/vectors) ===
+
=== Solution 2 (analytical/vectors) ===
 
Without loss of generality, place the pyramid in a 3-dimensional coordinate system such that <math>A = (1,0,0),</math> <math>B = (0,1,0),</math> <math>C = ( - 1,0,0),</math> <math>D = (0, - 1,0),</math> and <math>O = (0,0,z),</math> where <math>z</math> is unknown.
 
Without loss of generality, place the pyramid in a 3-dimensional coordinate system such that <math>A = (1,0,0),</math> <math>B = (0,1,0),</math> <math>C = ( - 1,0,0),</math> <math>D = (0, - 1,0),</math> and <math>O = (0,0,z),</math> where <math>z</math> is unknown.
  
Line 32: Line 32:
 
<cmath>z^{2}\sqrt {2} = 1 + z^{2}\implies z^{2} = 1 + \sqrt {2}.</cmath>
 
<cmath>z^{2}\sqrt {2} = 1 + z^{2}\implies z^{2} = 1 + \sqrt {2}.</cmath>
  
−
Now let's find <math>\cos \theta.</math> Let <math>\vec{u}</math> and <math>\vec{v}</math> be normal vectors to the planes containing faces <math>OAB</math> and <math>OBC,</math> respectively. It follows that letting
+
Now let's find <math>\cos \theta.</math> Let <math>\vec{u}</math> and <math>\vec{v}</math> be normal vectors to the planes containing faces <math>OAB</math> and <math>OBC,</math> respectively. From the definition of the [[dot product]] as <math>\vec{u}\cdot \vec{v} = \parallel \vec{u}\parallel \parallel \vec{v}\parallel \cos \theta</math>, we will be able to solve for <math>\cos \theta.</math> A cross product yields (alternatively, it is simple to find the equation of the planes <math>OAB</math> and <math>OAC</math>, and then to find their normal vectors)
−
 
 
−
<cmath>\vec{u}\cdot \vec{v} = \parallel \vec{u}\parallel \parallel \vec{v}\parallel \cos \theta</cmath>
 
−
 
 
−
will allow us to solve for <math>\cos \theta.</math> A cross product yields
 
  
 
<cmath>\vec{u} = \overrightarrow{OA}\times \overrightarrow{OB} = \left| \begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\
 
<cmath>\vec{u} = \overrightarrow{OA}\times \overrightarrow{OB} = \left| \begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\
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Hence, taking the dot product of <math>\vec{u}</math> and <math>\vec{v}</math> yields
 
Hence, taking the dot product of <math>\vec{u}</math> and <math>\vec{v}</math> yields
  
−
<cmath>- z^{2} + z^{2} + 1 = 1 = (\sqrt {1 + 2z^{2}})^{2}\cos \theta.</cmath>
+
<cmath>\cos \theta = \frac{ \vec{u} \cdot \vec{v} }{ \parallel \vec{u} \parallel \parallel \vec{v} \parallel } = \frac{- z^{2} + z^{2} + 1}{(\sqrt {1 + 2z^{2}})^{2}} = \frac {1}{3 + 2\sqrt {2}} = 3 - 2\sqrt {2} = 3 - \sqrt {8}.</cmath>
−
 
 
−
Simplifying,
 
−
 
 
−
<cmath>\cos \theta = \frac {1}{3 + 2\sqrt {2}} = 3 - 2\sqrt {2} = 3 - \sqrt {8}.</cmath>
 
  
 
Flipping the signs (we found the cosine of the supplement angle) yields <math>\cos \theta = - 3 + \sqrt {8},</math> so the answer is <math>\boxed{005}</math>.
 
Flipping the signs (we found the cosine of the supplement angle) yields <math>\cos \theta = - 3 + \sqrt {8},</math> so the answer is <math>\boxed{005}</math>.

Revision as of 13:20, 31 July 2008

Problem

Pyramid $OABCD$ has square base $ABCD,$ congruent edges $\overline{OA}, \overline{OB}, \overline{OC},$ and $\overline{OD},$ and $\angle AOB=45^\circ.$ Let $\theta$ be the measure of the dihedral angle formed by faces $OAB$ and $OBC.$ Given that $\cos \theta=m+\sqrt{n},$ where $m_{}$ and $n_{}$ are integers, find $m+n.$

Solution

Solution 1 (trigonometry)

size(220); defaultpen(linewidth(0.7)); currentprojection = perspective(5,3,2);
import three;
triple A = (1,0,0), B=(0,0,0), C=(0,1,0), D=(1,1,0), O=(1,1,(1+2^.5)^.5)/2^.5, P=O*(1-.5^.5); /* , P = foot(A, O, B) */
draw(A--B--C--D--A--O--B--O--C--O--D); draw(A--C,linewidth(0.6)); D(A--P--C); MP("A",A);MP("B",B);MP("C",C);MP("D",D);MP("P",P,NW);MP("\theta",P,(0.5,-4));MP("45^{\circ}",O,(2,-6)); MP("O",O,N); 
draw(rightanglemark(C,P,O,2)); draw(anglemark(A,P,C,3)); draw(anglemark(D,O,C,3));
 (Error making remote request. Unknown error_msg)

The angle $\theta$ is the angle formed by two perpendiculars drawn to $BO$, one on the plane determined by $OAB$ and the other by $OBC$. Let the perpendiculars from $A$ and $C$ to $\overline{OB}$ meet $\overline{OB}$ at $P.$ Without loss of generality, let $AP = 1.$ It follows that $\triangle OPA$ is a $45-45-90$ right triangle, so $OP = AP = 1,$ $OB = OA = \sqrt {2},$ and $AB = \sqrt {4 - 2\sqrt {2}}.$ Therefore, $AC = \sqrt {8 - 4\sqrt {2}}.$

From the Law of Cosines, $AC^{2} = AP^{2} + PC^{2} - 2(AP)(PC)\cos \theta,$ so

\[8 - 4\sqrt {2} = 1 + 1 - 2\cos \theta \Longrightarrow \cos \theta = - 3 + 2\sqrt {2} = - 3 + \sqrt{8}.\]

Thus $m + n = \boxed{005}$.

Solution 2 (analytical/vectors)

Without loss of generality, place the pyramid in a 3-dimensional coordinate system such that $A = (1,0,0),$ $B = (0,1,0),$ $C = ( - 1,0,0),$ $D = (0, - 1,0),$ and $O = (0,0,z),$ where $z$ is unknown.

We first find $z.$ Note that

\[\overrightarrow{OA}\cdot \overrightarrow{OB} = \parallel \overrightarrow{OA}\parallel \parallel \overrightarrow{OB}\parallel \cos 45^\circ.\]

Since $\overrightarrow{OA} =\, <1,0, - z>$ and $\overrightarrow{OB} =\, <0,1, - z> ,$ this simplifies to

\[z^{2}\sqrt {2} = 1 + z^{2}\implies z^{2} = 1 + \sqrt {2}.\]

Now let's find $\cos \theta.$ Let $\vec{u}$ and $\vec{v}$ be normal vectors to the planes containing faces $OAB$ and $OBC,$ respectively. From the definition of the dot product as $\vec{u}\cdot \vec{v} = \parallel \vec{u}\parallel \parallel \vec{v}\parallel \cos \theta$, we will be able to solve for $\cos \theta.$ A cross product yields (alternatively, it is simple to find the equation of the planes $OAB$ and $OAC$, and then to find their normal vectors)

\[\vec{u} = \overrightarrow{OA}\times \overrightarrow{OB} = \left| \begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & - z \\ 0 & 1 & - z \end{array}\right| =\, < z,z,1 > .\]

Similarly,

\[\vec{v} = \overrightarrow{OB}\times \overrightarrow{OC} - \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & - z \\ - 1 & 0 & - z \end{array}\right| =\, < - z,z,1 > .\]

Hence, taking the dot product of $\vec{u}$ and $\vec{v}$ yields

\[\cos \theta = \frac{ \vec{u} \cdot \vec{v} }{ \parallel \vec{u} \parallel \parallel \vec{v} \parallel } = \frac{- z^{2} + z^{2} + 1}{(\sqrt {1 + 2z^{2}})^{2}} = \frac {1}{3 + 2\sqrt {2}} = 3 - 2\sqrt {2} = 3 - \sqrt {8}.\]

Flipping the signs (we found the cosine of the supplement angle) yields $\cos \theta = - 3 + \sqrt {8},$ so the answer is $\boxed{005}$.

See also

1995 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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